斯托克、沃森着《计量经济学》第二版答案.pdf

PART ONE Solutions to Exercises Chapter 2 Review of Probability ? ? Solutions to Exercises 1. (a) Probability distribution function for Y Outcome (number of heads) Y ? 0 Y ? 1 Y ? 2 probability 0.25 0.50 0.25 (b) Cumulative probability distribution function for Y Outcome (number of heads) Y ? 0 0 ? Y ? 1 1 ? Y ? 2 Y ? 2 Probability 0 0.25 0.75 1.0 (c) = ( )(00.25)(1 0.50)(2 0.25)1.00 Y E Y? Using Key Concept 2.3: 22 var( )() ( ) ,YE YE Y? and ? 2222 ()(00.25)(10.50)(20.25)1.50E Y so that 222 var( )() ( )1.50(1.00)0.50.YE YE Y? 2. We know from Table 2.2 that ? ?Pr(0)0 22,Y ? ?Pr(1)0 78,Y ? ?Pr(0)0 30,X ? ?Pr(1)0 70.X So (a) ( )0Pr(0)1 Pr(1) 00 221 0 780 78, ( )0Pr(0)1 Pr(1) 00 301 0 700 70 Y X E YYY E XXX ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (b) 22 22 22 22 22 22 () (00.70)Pr(0)(10.70)Pr(1) ( 0 70)0 300 300 700 21 () (00.78)Pr(0)(10.78)Pr(1) ( 0 78)0 220 220 780 1716 XX YY E X XX E Y YY ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 4 Stock/Watson - Introduction to Econometrics - Second Edition (c) Table 2.2 shows Pr(0,0)0 15,XY? ? Pr(0,1)0 15,XY? ? Pr(1,0)0 07,XY? ? Pr(1,1)0 63.XY? ? So cov( ,)()() (0-0.70)(0-0.78)Pr(0,0) (00 70)(10 78)Pr(01) (10 70)(00 78)Pr(10) (10 70)(10 78)Pr(11) ( 0 70)( 0 78)0 15( 0 70)0 220 15 0 30( 0 78)0 070 XYXY X YE XY XY XY XY XY ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?300 220 63 0 084, 0 084 ( ,)0 4425 0 21 0 1716 XY XY cor X Y ? ? ? ? ? ? ? ? ? ? ? ? ? 3. For the two new random variables 36WX? and 207 ,VY? we have: (a) ( )(207 )207 ( )207 0 7814 54, ()(36 )36 ( )36 0 707 2 E VEYE Y E WEXE X ? ? ? ? ? ? ? ? (b) 222 222 var(36 )636 0 217 56, var(207 )( 7)49 0 17168 4084 WX VY X Y ? ? ? ? ? ? ? ? ? (c) (36, 207 )6( 7)( ,)420 0843 528 3 528 (,)0 4425 7 56 8 4084 WV WV WV covXYcov X Y cor W V ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 4. (a) ? 333 ()0(1)1E Xppp (b) ?()0(1)1 kkk E Xppp (c) ?()0.3E X ? 22 var()() ()0.30.090.21XE XE X Thus, ?0.210.46.? To compute the skewness, use the formula from exercise 2.21: 3323 23 ()()3 () ( )2 ( ) 0.33 0.32 0.30.084 E XE XE XE XE X? ? ? ? Alternatively, ? 333 () = (10.3)0.3(00.3)0.70.084E X? Thus, skewness 333 () /.084/0.460.87.E X? Solutions to Exercises in Chapter 2 5 To compute the kurtosis, use the formula from exercise 2.21: 443224 234 ()()4 ( ) ()6 ( ) ()3 ( ) 0.34 0.36 0.33 0.30.0777 E XE XE XE XE XE XE X? ? ? ? ? Alternatively, ? 444 () = (10.3)0.3(00.3)0.70.0777E X? Thus, kurtosis is 444 () /= .0777/0.461.76E X? ? 5. Let X denote temperature in ?F and Y denote temperature in ?C. Recall that Y ? 0 when X ? 32 and Y ? 100 when X ? 212; this implies (100/180) (32) or 17.78(5/9).YXYX? ? Using Key Concept 2.3, ?70 F X ? implies that 17.78(5/9) 7021.11 C, Y ? ? and ? ?7 F X ? implies(5/9) 73.89 C. Y ? 6. The table shows that Pr(0,0)0 045,XY? ? Pr(0,1)0 709,XY? ? Pr(1,0)0 005,XY? ? Pr(1,1)0 241,XY? ? Pr(0)0 754,X? ? ? ?Pr(1)0 246,X ? ?Pr(0)0 050,Y ? ?Pr(1)0 950.Y (a) ? ? ? ? ? ? ? ( )0Pr(0)1 Pr(1) 00 0501 0 9500 950 Y E YYY? (b) (unemployed) Unemployment Rate (labor force) Pr(0)0 05010 9501( ) # # YE Y ? ? ? ? ? ? (c) Calculate the conditional probabilities first: Pr(0,0)0 045 Pr(0|0)0 0597, Pr(0)0 754 XY YX X ? ? ? ? Pr(0,1)0 709 Pr(1|0)0 9403, Pr(0)0 754 XY YX X ? ? ? ? Pr(1,0)0 005 Pr(0|1)0 0203, Pr(1)0 246 XY YX X ? ? ? ? Pr(1,1)0 241 Pr(1|1)0 9797 Pr(1)0 246 XY YX X ? ? ? ? The conditional expectations are ( |1)0Pr(0|1)1 Pr(1|1) 00 0203 1 0 97970 9797, ( |0)0Pr(0|0)1 Pr(1|0) 00 05971 0 94030 9403 E Y XYXYX E Y XYXYX ? ? ? ? ? ? ? ? ? ? ? ? ? ? 6 Stock/Watson - Introduction to Econometrics - Second Edition (d) Use the solution to part (b), Unemployment rate for college grads 1( |1)10.97970.0203. Unemployment rate for non-collegegrads 1( |0)10.94030.0597. E Y X E Y X ? ? ? ? ? ? (e) The probability that a randomly selected worker who is reported being unemployed is a college graduate is Pr(1,0)0 005 Pr(1|0)0 1 Pr(0)0 050 XY XY Y ? ? ? ? ? The probability that this worker is a non-college graduate is Pr(0|0)1Pr(1|0)10 10 9XYXY? ? ? ? ? ? ? (f) Educational achievement and employment status are not independent because they do not satisfy that, for all values of x and y, Pr(|)Pr()YyXxYy? For example, Pr(0|0)0 0597Pr(0)0 050YXY? ? ? 7. Using obvious notation, ?;CMF thus ? CMF ? and 222 2cov(,). CMF M F? This implies (a) ?4045$85,000 C ?per year. (b) (,) (,), MF Cov M F cor M F ? ? ? so that (,)(,). MF Cov M Fcor M F? ? Thus (,)12 18 0.80172.80,Cov M F ? where the units are squared thousands of dollars per year. (c) 222 2cov(,), CMF M F? so that 222 12182 172.80813.60, C ? ? and ?813.6028.524 C ? thousand dollars per year. (d) First you need to look up the current Euro/dollar exchange rate in the Wall Street Journal, the Federal Reserve web page, or other financial data outlet. Suppose that this exchange rate is e (say e ? 0.80 euros per dollar); each 1$ is therefore with eE. The mean is therefore e?C (in units of thousands of euros per year), and the standard deviation is e?C (in units of thousands of euros per year). The correlation is unit-free, and is unchanged. 8. ?( )1, Y E Y? 2 var( )4. Y Y? With ? 1 2( 1),ZY 22 111 (1)(1)(1 1)0, 222 111 var(1)41 244 ZY ZY EY Y ? ? ? ? ? ? ? ? ? ? ? Solutions to Exercises in Chapter 2 7 9. Value of Y 14 22 30 40 65 Probability Distribution of X 1 0.02 0.05 0.10 0.03 0.01 0.21 5 0.17 0.15 0.05 0.02 0.01 0.40 Value of X 8 0.02 0.03 0.15 0.10 0.09 0.39 Probability distribution of Y 0.21 0.23 0.30 0.15 0.11 1.00 (a) The probability distribution is given in the table above. 222222 22 ( )14 0.2122 0.23300.30400.1565 0.1130.15 ()140.21220.23300.30400.15650.111127.23 Var(Y)() ( )218.21 14.77 Y E Y E Y E YE Y ? ? ? ? ? (b) Conditional Probability of Y|X ? 8 is given in the table below Value of Y 14 22 30 40 65 0.02/0.39 0.03/0.39 0.15/0.39 0.10/0.39 0.09/0.39 ( |8)14 (0.02/0.39)22 (0.03/0.39)30 (0.15/0.39) 40 (0.10/0.39)65 (0.09/0.39)39.21 E Y X? ? 2222 22 (|8)14(0.02/0.39)22(0.03/0.39)30(0.15/0.39) 40(0.10/0.39)65(0.09/0.39)1778.7 E Y X ? ? 2 8 Var( )1778.739.21241.65 15.54 Y X Y ? ? ? ? ? (c) () (1 14 0.02)(1 22:0.05)(8 65 0.09)171.7E XY?L Cov( , )()( ) ( )171.75.33 30.1511.0X YE XYE X E Y? Corr( , )Cov( , )/()11.0/(5.46 14.77)0.136 XY X YX Y? ? 10. Using the fact that if 2 , YY YN? ? ? ? : then (0,1) Y Y Y N ? ? ? and Appendix Table 1, we have (a) 131 Pr(3)Pr(1)0 8413 22 Y Y ? ? ? ? ? ? (b) Pr(0)1Pr(0) 303 1Pr1( 1)(1)0 8413 33 YY Y ? ? ? ? ? ? ? ? ? ? ? 8 Stock/Watson - Introduction to Econometrics - Second Edition (c) 4050505250 Pr(4052)Pr 555 (0 4)( 2)(0 4)1(2) 0 655410 97720 6326 Y Y ? ? ? ? ? ? ? ? ? ? ? ? ? (d) 65585 Pr(68)Pr 222 (2 1213)(0 7071) 0 98310 76020 2229 Y Y ? ? ? ? ? ? ? ? ? ? 11. (a) 0.90 (b) 0.05 (c) 0.05 (d) When 2 10 ,Y? then 10, /10 .YF ? (e) 2, YZ? where N(0,1),Z thus Pr(1)Pr( 11)0.32.YZ? ? 12. (a) 0.05 (b) 0.950 (c) 0.953 (d) The tdf distribution and N(0, 1) are approximately the same when df is large. (e) 0.10 (f) 0.01 13. (a) 2222 ()Var( )101;()Var()1000100. YW E YYE WW? ? (b) Y and W are symmetric around 0, thus skewness is equal to 0; because their mean is zero, this means that the third moment is zero. (c) The kurtosis of the normal is 3, so 4 $ () 3; Y Y E Y? ? ? ? solving yields 4 E()3;Y? a similar calculation yields the results for W. (d) First, condition on 0,X ? so that :SW? 2342 . ( |0)0;(|0)100,(|0)0,(|0)3 100E SXE S XE S XE S X? ? Similarly, 234 ( |1)0;(|1)1,(|1)0,(|1)3.E SXE S XE S XE S X? From the large of iterated expectations ( )( |0)Pr(X0)( |1)Pr(1)0E SE S XE S XX? 222 ()(|0) Pr(X0)(|1) Pr(1)1000.01 1 0.991.99E SE S XE S XX? ? 333 ()(|0)Pr(X0)(|1)Pr(1)0E SE S XE S XX? 4442 ()(|0) Pr(X0)(|1) Pr(1)3 1000.013 1 0.99302.97E SE S XE S XX? ? ? ? Solutions to Exercises in Chapter 2 9 (e) ( )0, S E S? thus 33 ()()0 S E SE S?from part d. Thus skewness ? 0. Similarly, 222 ()()1.99, SS E SE S? and 44 ()()302.97. S E SE S? Thus, 2 kurtosis302.97/(1.99 )76.5? 14. The central limit theorem suggests that when the sample size (n) is large, the distribution of the sample average ( )Y is approximately 2 , YY N? ? ? ? with 2 2 . Y nY ? ? Given 100, Y ? 2 43 0, Y ? (a) 100,n ? 2 2 43 100 0 43, Y nY ? ? ? and 100101 100 Pr(101)Pr(1 525)0 9364 0 430 43 Y Y ? ? ? ? ? ? ? (b) 165,n ? 2 2 43 165 0 2606, Y nY ? ? ? and 10098 100 Pr(98)1Pr(98)1Pr 0 26060 2606 1( 3 9178)(3 9178)1 000 (rounded to four decimal places) Y YY ? ? ? ? ? ? ? ? ? ? ? ? ? (c) 64,n ? 2 2 43 6464 0 6719, Y Y ? ? ? and 101 100100103 100 Pr(101103)Pr 0 67190 67190 6719 (3 6599)(1 2200)0 99990 88880 1111 Y Y ? ? ? ? ? ? ? ? ? ? 15. (a) 9.6 101010.4 10 Pr(9.610.4)Pr 4/4/4/ 9.6 1010.4 10 Pr 4/4/ Y Y nnn Z nn ? ? ? ? ? ? ? ? ? ? where Z N(0, 1). Thus, (i) n ? 20; 9.6 1010.4 10 PrPr( 0.890.89)0.63 4/4/ ZZ nn ? ? ? ? ? (ii) n ? 100; 9.6 1010.4 10 PrPr( 2.002.00)0.954 4/4/ ZZ nn ? ? ? ? ? (iii) n ? 1000; 9.6 1010.4 10 PrPr( 6.326.32)1.000 4/4/ ZZ nn ? ? ? ? ? 10 Stock/Watson - Introduction to Econometrics - Second Edition (b) 10 Pr(1010)Pr 4/4/4/ Pr. 4/4/ cYc cYc nnn cc Z nn ? ? ? ? ? ? ? ? ? ? As n get large 4/ c n gets large, and the probability converges to 1. (c) This follows from (b) and the definition of convergence in probability given in Key Concept 2.6. 16. There are several ways to do this. Here is one way. Generate n draws of Y, Y1, Y2, Yn. Let Xi ? 1 if Yi ? 3.6, otherwise set Xi ? 0. Notice that Xi is a Bernoulli random variables with ?X ? Pr(X ? 1) ? Pr(Y ? 3.6). Compute .X Because X converges in probability to ?X ? Pr(X ? 1) ? Pr(Y ? 3.6), X will be an accurate approximation if n is large. 17. ?Y ? 0.4 and 2 4 Y ? (a) (i) P(Y ? 0.43) ? 0.40.430.40.4 PrPr0.61240.27 0.24/0.24/0.24/ YY nnn ? ? ? ? (ii) P(Y ? 0.37) ? 0.40.370.40.4 PrPr1.220.11 0.24/0.24/0.24/ YY nnn ? ? ? ? ? (b) We know Pr(?1.96 ? Z ? 1.96) ? 0.95, thus we want n to satisfy 0.41 0.4 0.24/ 0.411.96 n ? ? ? and 0.39 0.4 0.24/ 1.96. n ? ? ? Solving these inequalities yields n ? 9220. 18. Pr(0)0 95,Y$? ? Pr(20000)0 05.Y$? ? (a) The mean of Y is 0Pr(0)20,000Pr(20000)1000. Y Y$Y$? The variance of Y is ? ? 2 2 2 2 227 Pr0(200001000)Pr(20000) (01000) ( 1000)0 95 190000 051 9 10 , YY E Y YY ? ? ? ? ? ? ? ? ? ? ? ? ? ? so the standard deviation of Y is 1 2 7 (1 9 10 )4359 Y $? ? (b) (i) ( )1000, Y E Y$? 2 7 25 1 9 10 100 1 9 10 . Y nY ? ? ? ? ? ? ? (ii) Using the central limit theorem, 55 Pr(2000)1Pr(2000) 10002 0001 000 1Pr 1 9 101 9 10 1(2 2942)10 98910 0109 YY Y ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? Solutions to Exercises in Chapter 2 11 19. (a) 1 1 Pr()Pr(,) Pr(|)Pr() l jij i l jii i YyXx Yy Y y X xX x ? ? ? ? ? ? (b) 111 1 1 1 ( )Pr()Pr(|)Pr() Pr(|) Pr() ( |)Pr() i ii kkl jjjjii jji k l jji j i l i E YyYyyYy XxXx yYy XxX x E Y X xX x ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (c) When X and Y are independent, Pr(,)Pr()Pr() ijij Xx YyXxYy? so 11 11 11 ()() ()()Pr(,) ()()Pr()Pr() ()Pr()()Pr( () ()000, XYXY lk iXjYij ij lk iXjYij ij lk iXijYj ij XY E XY xyX x Y y xyX xY y xXxyYy E XE Y ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 0 ( ,)0 XY XYXY cor X Y ? ? ? ? ? ? 20. (a) 11 Pr()Pr(|,)Pr(,) lm iijhjh jh YyYy XxZzXxZz ? ? ? (b) 1 111 111 11 ( )Pr()Pr() Pr(|,)Pr(,) Pr(|,) Pr(,) ( |,)Pr(,) k iii i klm iijhjh ijh lmk iijhjh jhi lm jhjh jh E YyYyYy yYy XxZzXxZz yYy XxZzXxZz E Y XxZzXxZz ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 12 Stock/Watson - Introduction to Econometrics - Second Edition where the first line in the definition of the mean, the second uses (a), the third is a rearrangement, and the final line uses the definition of the conditional expectation. 21. (a) 32322223 32233223 323 ()() ()22 ()3 ()3 ()()3 () ()3 () () () ()3 () ()2 () E XE XXE XXXXX E XE XE XE XE XE XE XE XE X E XE XE XE X ? ? ? ? ? (b) 43223 4322332234 432234 43224 ()(33)() 3333 ()4 () ()6 () ()4 () ()() ()4 () ()6 () ()3 () E XE XXXX E XXXXXXX E XE XE XE XE XE X E XE X E XE XE XE XE XE X ? ? ? ? ? ? 22. The mean and variance of R are given by 2222 0.08(1)0.05 0.07(1)0.0422(1) 0.070.040.25 ww wwww ? ? ? ? where 0.070.04 0.25(, ) sb Cov RR? follows from the definition of the correlation between Rs and Rb. (a) 0.065;0.044? (b) 0.0725;0.056? (c) w ? 1 maximizes ;0.07? ? for this value of w. (d) The derivative of ?2 with respect to w is 2 22 2.072(1) 0.04(24 ) 0.07 0.04 0.25 0.01020.0018 d www dw w ? ? ? solving for w yields 18/1020.18.w ?(Notice that the second derivative is positive, so that this is the global minimum.) With 0.18,.038. R w? 23. X and Z are two independently distributed standard normal random variables, so 22 0,1,0. XZXZXZ ? (a) Because of the independence between X and ,Z Pr(|)Pr(),ZzXxZz? and ( | )( )0.E Z XE Z? Thus 2222 ( | )(| )(| )( | )0E Y XE XZ XE X XE Z XXX? (b) 222 ()1, XX E X? and 22 ()()101 YZ E XZE X? ? ? (c) 33 ()()()().E XYE XZXE XE ZX? Using the fact that the odd moments of a standard normal random variable are all zero, we have 3 ()0.E X? Using the independence between X and ,Z we have ()0. ZX E ZX? ? Thus 3 ()()()0.E XYE XE ZX? Solutions to Exercises in Chapter 2 13 (d) ()()()(0)(1) ()()() 000 0 ( ,)0 XY XY XYXY Cov XYE XYE XY E XYXE XYE X cor X Y ? ? ? ? ? ? ? ? ? ? ? 24. (a) 2222 () i E Y? and the result follows directly. (b) (Yi/?) is distributed i.i.d. N(0,1), 2 1( / ) , n i i WY? ? ? and the result follows from the definition of a 2 n ? random variable. (c) E(W) ? 22 22 11 (). nn ii ii YY E WEEn ? ? ? ? (d) Write 22 22 11 ( / ) 11 / nn iii YY nn YY V ? ? ? ? ? ? which follows from dividing the numerator and denominator by ?. Y1/? N(0,1), 2 2( / ) n i i Y? ? ? 2 1n ? ? , and Y1/? and 2 2( / ) n i i Y? ? ? are independent. The result then follows from the definition of the t distribution. Chapter 3 Review of Statistics ? ? Solutions to Exercises 1. The central limit theorem suggests that when the sample size (n) is large, the distribution of the sample average (Y) is approximately 2 , YY N? ? ? ? with 2 2 . Y nY ? ? Given a population 100, Y ? 2 43 0, Y ? we have (a) 100,n? 2 2 43 100 0 43, Y nY ? ? ? and 100101 100 Pr(101)Pr(1.525)0 9364 0 430 43 Y Y ? ? ? ? ? ? ? (b) 64,n ? 2 2 43 6464 0 6719, Y Y ? ? ? and 101 100100103 100 Pr(101103)Pr 0 67190 67190 6719 (3 6599)(1 2200)0 99990 88880 1111 Y Y ? ? ? ? ? ? ? ? ? ? ? (c) 165,n ? 2 2 43 165 0 2606, Y nY ? ? ? and 10098 100 Pr(98)1Pr(98)1Pr 0 26060 2606 1( 3 9178)(3 9178)1 0000 (rounded to four decimal places) Y YY ? ? ? ? ? ? ? ? ? ? ? ? ? 2. Each random draw i Y from the Bernoulli distribution takes a value of either zero or one with probability Pr(1) i Yp? and Pr(0)1. i Yp? ? The random variable i Y has mean ( )0Pr(0)1 Pr(1), i E YYYp? ? and variance 2 22 22 var( )() (0)Pr(0)(1)Pr(1) (1)(1)(1) iiY YE Y pYpY ii ppp ppp ? ? ? Solutions to Exercises in Chapter 3 15 (a) The fraction of successes is 1 (1)(success) n iii Y# Y# pY nnn ? ? ? (b) 1 11 11 ( )( ) n nn ii i ii Y E pEE Ypp nnn ? ? ? ? ? ? ? ? (c) 1 22 11 11(1) var( )varvar( )(1) n nn ii i ii Ypp pYpp nnnn ? ? ? ? ? ? ? ? ? The second equality uses the fact that 1 Y, ?, Yn are i.i.d. draws and cov( ,)0, ij Y Y? for .ij? 3. Denote each voters preference by .Y 1Y ? if the voter prefers the incumbent and 0Y ? if the voter prefers the challenger. Yis a Bernoulli random variable with probability Pr(1)Yp? and Pr(0)1.Yp? ? From the solution to Exercise 3.2, Y has mean p and variance (1).pp? (a) 215 400 0 5375.p ? ? (b) (1)0.5375(10.5375) 4 400 var( )6 2148 10 . pp n p ? ? ? The standard error is SE 1 2 ( )(var( )0 0249.pp? ? (c) The computed t-statistic is 0 0 53750 5 1 506 SE( )0 0249 p act p t p ? ? ? ? ? ? ? ? Because of the large sample size (400),n ? we can use Equation (3.14) in the text to get the p-value for the test 0 0 5Hp? ? vs. 1 0 5:Hp? ? -value2 ( |)2 ( 1 506)2 0 0660 132 act pt? ? ? ? ? ? ? ? (d) Using Equation (3.17) in the text, the p-value for the test 0 0 5Hp? ? vs. 1 0 5Hp? ? is -value1()1(1 506)10 9340 066 act pt? ? ? ? ? ? ? (e) Part (c) is a two-sided test and the p-value is the area in the tails of the standard normal distribution outside ? (calculated t-statistic). Part (d) is a one-sided test and the p-value is the area under the standard normal distribution to the right of the calculated t-statistic. (f) For the test 0 0 5Hp? ? vs. 1 0 5,Hp? ? we cannot reject the null hypothesis at the 5% significance level. The p-value 0.066 is larger than 0.05. Equivalently the calculated t-statistic 1 506? is less than the critical value 1.645 for a one-sided test with a 5% significance level. The test suggests that the survey did not contain statistically significant evidence that the incumbent was ahead of the challenger at the time of the survey. 16 Stock/Watson - Introduction to Econometrics - Second Edition 4. Using Key Concept 3.7 in the text (a) 95% confidence interval for p is 1.96( )0.5375 1.96 0.0249(0.4887,0.5863).pSE p? (b) 99% confidence interval for p is 2.57( )0.53752.57 0.0249(0.4735,0.6015).pSE p? (c) The interval in (b) is wider because of a larger